In (1​R,2​R,3​S,4​S)-1,3-dichloro-2,4-dimethylcyclobutane, are the blue lines not planes of symmetry?


My book says it has a centre of symmetry, but not a plane of symmetry. But if I took a cut through any of the blue lines wouldn't both sides be symmetrical?

  • 3
    $\begingroup$ One of your suggested planes transforms upward-facing chlorine to downward-facing chlorine and vice versa (which is not the behavior expected of a vertical plane of symmetry). Another does the same to methyl groups. $\endgroup$ Commented Jul 26, 2016 at 6:52
  • 1
    $\begingroup$ The point group is C$_i$ with only a centre of inversion (and the identity). This can also be described as an S$_2$ operation which is rotation by 180$^o$ and reflection in the plane perpendicular to the rotation axis. This axis points out of the plane of the carbons. $\endgroup$
    – porphyrin
    Commented Jul 26, 2016 at 8:41

1 Answer 1


Ivan already stated quite briefly in the comments, that the planes you drew would transform an upward facing chlorine to a downward facing chlorine and vice versa, and that the same holds for the methyl groups. This is of course true and the first and most obvious point, why there is no symmetry plane.

A more close investigation will, however, point out another important feature of the compound itself. The cyclobutane ring itself is not planar. In the (electronically) lowest conformation of the molecule it has no symmetry whatsoever (technically $C_1$ symmetry), not even inversion symmetry like your book states. This is due to the fact, that the four-membered ring is puckered. This is best visualised by looking at the 3D structure of the compound. There are multiple programs that will allow you to view such a structure and I will xyz coordinates at the end of this posting.

The following image will hopefully already show you what I am talking about. It was optimised on the DF-BP86/def2-SVP level of theory.

dihedral angle of the cyclobutane ring

In order for the the molecule to have inversion symmetry, the shown angle needs to be exactly $0^\circ$. The following animation shows the molecule rotating hopefully further clarifying this point.

molecule rotating

You can clearly see the puckering of the cyclobutane ring.

One final comment: For a molecule like this it really is necessary to specify the stereochemistry with the name. There are other stereoisomers, which do indeed have a symmetry plane. I leave it to you to figure out which of them do.

stereoisomers of 1,3-dichloro-2,4-dimethylcyclobutane

Except the last one, all of them have at least one plane of symmetry.

Addendum: Transition state structure

The symmetric ($C_\mathrm{i}$) transition state is on the DF-BP86/def2-SVP level of theory only about $\Delta G (298.15~\mathrm{K}) = 8~\mathrm{kJ/mol}$ higher in energy than the puckered asymmetric forms. It is therefore worthwhile to have a quick look at its structure. In the following graphics the red ball indicates the inversion centre.
Below that is an animation of the transitional mode. It briefly pauses at the puckered structures (these are extrapolated, not optimised structures) and the transition state to make it easier to see the movement in between them.
I would expect this process to be readily available at room temperature and therefore almost impossible to distinguish the different lower lying structures.

symmetric transition state
transitional mode

Coordinates of the lowest structure:

G09D.01; E(RB-P86/def2-SVP/W06) = -1154.66760674 A.U.
C        0.051334000      1.246599000      0.089432000
C       -0.768087000      0.117382000      0.789824000
C        0.002884000     -0.950933000     -0.020424000
C        1.086640000      0.142749000     -0.293350000
H        0.449081000      2.008047000      0.790972000
H       -0.535742000      0.060683000      1.871003000
H       -0.532992000     -1.105843000     -0.982729000
H        1.517474000      0.183811000     -1.312785000
C        0.343625000     -2.291820000      0.610089000
H        0.939420000     -2.156389000      1.535300000
H       -0.586821000     -2.839917000      0.866159000
H        0.932752000     -2.926935000     -0.083624000
C       -0.606103000      1.926898000     -1.110977000
H       -1.475939000      2.532182000     -0.784546000
H        0.116584000      2.600709000     -1.616245000
H       -0.977177000      1.195914000     -1.858831000
Cl       2.502295000      0.083512000      0.845211000
Cl      -2.570729000      0.126790000      0.662589000

Coordinates of the transition state (X is a dummy atom indicating the inversion centre)

G09D.01; E(RB-P86/def2-SVP/W06) = -1154.6659061600 A.U.
C       -0.188954000      1.025076554      0.415795307
C       -0.873066584     -0.383572154      0.501227861
C        0.188954000     -0.991627446     -0.457606693
C        0.898153416      0.404477846     -0.509590139
H        0.242239584      1.289691385      1.404629307
H       -0.864241753     -0.820445015      1.518694431
H       -0.267326416     -1.206068615     -1.446440693
H        0.939502247      0.849712985     -1.522875569
C        0.953473154     -2.212350139      0.042720768
H        1.426842139     -2.018382400      1.026791199
H        0.267219307     -3.079692431      0.145048491
H        1.760843738     -2.494108569     -0.663022216
C       -0.974378846      2.203987861     -0.147249232
H       -1.806836262      2.489927431      0.529225784
H       -0.309030693      3.083873569     -0.266301509
H       -1.418479861      1.964027600     -1.135500801
Cl       2.591747000      0.489781723      0.126797539
Cl      -2.591747000     -0.506506277     -0.064080461
X        0.006271708      0.013588700     -0.012543416
  • $\begingroup$ The model seen in your animation is even chiral. One notices that the two Cl-C bonds (green/black sticks) have different (non-oriented) angles to the axis of rotation you have chosen. If for example, I "hold" the animation at the "stop" where the upper methyl group is closest to my eyes, I can see that the Cl-C bond then to the right is more "vertical" (smaller angle with the axis you chose for the rotation), and the Cl-C bond to the left is more "horizontal". If you had given us the mirror image, that would be opposite. So the model you present, is chiral. $\endgroup$ Commented Jul 26, 2016 at 14:14
  • 1
    $\begingroup$ @Jeppe I was trying to imply that with stating that it has no symmetry (C1). The transformation of one of the forms into the other at room temperature will be very fast at room temperature and I would not be surprised if you find no chirality since the transition state does have an inversion centre and is therefore achiral. $\endgroup$ Commented Jul 27, 2016 at 6:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.